Problem 13 - College Algebra

Question

Solve each system by substitution. $$ \begin{array}{l} 4 x+2 y=-10 \\ 3 x+9 y=0 \end{array} $$

Step-by-Step Solution

Answer

The solution to the system is \((-3, 1)\).

Step 1: Solve for One Variable

Start by choosing one of the equations and solving for one of the variables. Let's solve the first equation for \( x \). The equation is \( 4x + 2y = -10 \). Isolate \( x \) by subtracting \( 2y \) from both sides: \[ 4x = -10 - 2y \] Now, divide every term by 4: \[ x = \frac{-10 - 2y}{4} = \frac{-5}{2} - \frac{y}{2} \].

Step 2: Substitute in the Second Equation

Substitute the expression for \( x \) from Step 1 into the second equation, \( 3x + 9y = 0 \). So, we substitute \( x = \frac{-5}{2} - \frac{y}{2} \) into the equation: \[ 3\left(\frac{-5}{2} - \frac{y}{2}\right) + 9y = 0 \].

Step 3: Simplify and Solve for y

Distribute the 3 across the terms inside the parenthesis: \[ \frac{-15}{2} - \frac{3y}{2} + 9y = 0 \]. Convert \( 9y \) to have a common denominator: \[ \frac{-15}{2} - \frac{3y}{2} + \frac{18y}{2} = 0 \]. Combine like terms: \[ \frac{-15 + 15y}{2} = 0 \]. Multiply the entire equation by 2 to eliminate the denominator: \[ -15 + 15y = 0 \]. Add 15 to both sides: \[ 15y = 15 \]. Divide both sides by 15: \[ y = 1 \].

Step 4: Substitute y Back to Find x

Now that we know \( y = 1 \), substitute it back into the expression we have for \( x \): \[ x = \frac{-5}{2} - \frac{1}{2}(1) = \frac{-5}{2} - \frac{1}{2} = \frac{-6}{2} = -3 \].

Step 5: Write the Solution

The values of \( x \) and \( y \) that solve the system of equations are \( x = -3 \) and \( y = 1 \). Therefore, the solution to the system is the ordered pair \((-3, 1)\).

Key Concepts

Substitution Method

The substitution method is a powerful technique for solving systems of equations. It involves replacing one variable with an expression that represents it using another equation. This method simplifies the process of finding solutions by reducing a system of equations into a single equation.

How It Works

• Choose one equation and solve for one variable in terms of the other(s).
• Substitute this expression into the other equation(s).
• Solve the new equation for the remaining variable.
• Substitute back to find the value of the first variable.

This step-by-step approach makes it easier to handle even complex systems through simplification and gradual substitution.

Linear Equations

Linear equations are equations of the first degree, meaning they contain variables raised only to the power of one. In a linear equation, each term is either a constant or the product of a constant and a single variable.

Characteristics

• Typically appear in the format: ax + by = c.
• When graphed, they create a straight line.
• The solutions correspond to intersections on a graph.

Understanding linear equations is essential because they are foundational in algebra and appear frequently in various applied mathematics problems.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to isolate variables and solve for them. This is a crucial skill to effectively solve equations, as demonstrated in the substitution method.

Steps Involved

• Use basic operations like addition, subtraction, multiplying, or dividing both sides of the equation.
• Apply distributive properties to expand expressions where necessary.
• Combine like terms to simplify the expressions.
• Reverse operations to isolate variables.

Mastery of algebraic manipulation allows for more efficient problem-solving and the ability to tackle more complex algebraic challenges.

Ordered Pairs

Ordered pairs are used to represent solutions of systems of equations, particularly when dealing with two variables. An ordered pair takes the form \( (x, y) \) and signifies a specific point on a graph.

Understanding Ordered Pairs

• The first number (x-coordinate) indicates horizontal positioning.
• The second number (y-coordinate) indicates vertical positioning.
• Ordered pairs make it easy to cross-reference solutions with graphical representations.

In the context of solving systems, identifying the ordered pair from equations confirms the solution by checking it against the system’s constraints.